Answer:
The energy stored in capacitor B is four times the energy stored in capacitor A
Explanation:
First let the write-out the formula for the energy stored in a capacitor in-terms of the capacitance(C)and the voltage(V)
[tex]\\W_{energy}=\frac{1}{2}CV^{2}\\[/tex]
For capacitor A charged to a voltage of V, The energy stored is
[tex]\\W_{A}=\frac{1}{2}CV^{2}\\[/tex]
and for capacitor B, we have [tex]\\W_{B}=\frac{1}{2}C[2V]^{2}\\[/tex]
[tex]\\ W_{B}=2CV^{2}\\[/tex]
From the energy stored in Capacitor A we can arrive at
[tex]2W_{A}=CV^{2}\\[/tex]
if we substitute this value into the energy for capacitor B we arrive at
[tex]\\W_{B}=2(2W_{A} )\\W_{B}=4W_{A}\\[/tex]
Hence we can conclude that the energy stored in capacitor B is four times the energy stored in capacitor A
Answer:
The stored energy (Potential difference) of capacitor B which is equal to 2V is twice as much as capacitor A which has a stored capacity of 1V or V. Increased voltage also implies a reduction in the current passing through it.
Explanation:
Potential Difference is simply the work that has to be done in transferring a unit positive charge from one point to the other. It is also referred to as the "Electric Potential".
Also, the potential difference between points A and B, V(B) - V(A), is the change in Potential energy of a charge (q) moved from A to B, divided by the charge. Its units are joules per coulomb which is also equal to volt (V).
Note: the relationship between Potential energy and Potential difference is expressed thus: Change in Potential Difference = Change in Potential Energy / Charge.
